This table represents a quadratic function with a vertex at [tex]$(1,2)$[/tex]. What is the average rate of change for the interval from [tex]$x=5$[/tex] to [tex]$x=6$[/tex]?

\begin{tabular}{|l|l|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
1 & 2 \\
\hline
2 & 3 \\
\hline
3 & 6 \\
\hline
4 & 11 \\
\hline
5 & 18 \\
\hline
\end{tabular}

A. 27
B. 3
C. 18
D. 9



Answer :

To find the average rate of change of the quadratic function on the interval from [tex]\(x = 5\)[/tex] to [tex]\(x = 6\)[/tex], we proceed as follows:

1. Identify the function values at the given points:
- For [tex]\(x = 5\)[/tex], the table shows [tex]\(y = 18\)[/tex].
- For [tex]\(x = 6\)[/tex], based on the quadratic function given and its vertex at [tex]\((1,2)\)[/tex], the value is [tex]\(y = 27\)[/tex].

2. Determine the interval length:
- The interval length [tex]\(\Delta x\)[/tex] is [tex]\(6 - 5 = 1\)[/tex].

3. Calculate the average rate of change:
- The average rate of change is given by the formula:
[tex]\[ \text{Average Rate of Change} = \frac{\Delta y}{\Delta x} = \frac{y(6) - y(5)}{6 - 5} \][/tex]
- Substituting the known values:
[tex]\[ \text{Average Rate of Change} = \frac{27 - 18}{1} = \frac{9}{1} = 9 \][/tex]

Therefore, the average rate of change for the interval from [tex]\(x=5\)[/tex] to [tex]\(x=6\)[/tex] is [tex]\(9\)[/tex].

So the correct answer is:

D. 9